\section{Limit behavior of the regulator and linear independence of the elements}
In this section, we will establish a limit formula of the regulator of $M_i$ constructed in the last section. First we transform the curve to a family with one parameter which goes to 0, then we give a basis of $H_1(X;\mathbb{Z})^-$ of the curve when the parameter is close to zero, with this basis we show the limit of the regulator of $M_i$ does not vanish. So we can conclude that the elements are linearly independent for a general curve in the family.

Let $x \rightarrow \frac{1}{x}, y\rightarrow \frac{\lambda (y-1) x \prod_{i=1}^g(x+a_i) + 2}{2x^{g+1}}$, then (\ref{eqn:curve}) becomes
\begin{equation}\label{eqn:prodcurve}
\lambda(y-1)(y+1)x\prod_{i=1}^g (x+a_i)+4y=0.
\end{equation}
Then under the transformation

 \begin{eqnarray}
 M_i &= & \left\{\frac{4}{(\lambda (y-1) x \prod_{i=1}^g(x+a_i) + 2)^2}, a_i \frac{1}{x}+1\right\} \nonumber \\
 &=& \left\{\frac{4}{-(\frac{4y}{y+1} + 2)^2} ,  \frac{x+a_i}{x} \right\} \nonumber \\
 &=& \left\{\left(\frac{y+1}{y-1}\right)^2 ,  \frac{x+a_i}{x} \right\}.  \label{eqn:element}
 \end{eqnarray}



We can see the curve in (\ref{eqn:prodcurve}) as a family in the parameter $t = 1/\lambda$ and with $a_i, i=1,\ldots,g$ fixed, we denote the projective model of this family as $\pi: Y \rightarrow \mathbb{P}^1$. Let $D$ be a small disk around $t=0$ and $D^*$ be the punctured disk $D\backslash \{t=0\}$. The fiber at any $t \in D^*$ is a smooth projective curve of genus $g$ and the fiber at $t=0$ denoted by $Y_0$ is a bunch of lines looks like Figure (\ref{fig:badfiber}).

\setlength{\unitlength}{1mm}
\begin{figure}[h]
\centering
\begin{picture}(60,20)
% lines
\multiput(-5, 5)(0, 10){2}{\line(1, 0){60}}
\multiput(0, 0)(10, 0){6}{\line(0, 1){20}}
% dots for homology base
% dots
\multiput(10, 15)(10, 0){5}{\circle*{2}}
\end{picture}
\caption{Configuration of Bad Fiber}
\label{fig:badfiber}
\end{figure}

\begin{lemma}\label{lemma:limit}
Let $\pi:Y \rightarrow \mathbb{P}^1$ be a algebraic surface defined by $f(x,y)xy=t$ around $x=y=0$ with parameter $t$ and $f(0,0) \neq 0$, $p : Y \rightarrow \mathbb{A}^1$ be the projection to the coordinate x. Let $\gamma_t$ be the lift of a counterclockwise simple loop around zero on the x-axis seen as a complex plane to the fiber of $\pi$ above $t$ such that $\gamma_t$ is continuous in $t$ and $\gamma_0$ is a counterclockwise simple loop on $y=0$ as $t=0$. Denote $\eta(t)=\int_{\gamma_t} \eta(\{u(x,y)x^a,v(x,y)y^b\})$ as a function of $t$ where $u,v$ are holomorphic functions on the surface which does not vanish at $x=y=0$. Then for real $t$
\begin{equation*}
\lim_{t\rightarrow 0} \frac{\eta(t)}{\log \abs{t}} = 2ab\pi
\end{equation*}

\end{lemma}
\begin{proof}
We prove $\lim_{t\rightarrow 0} t\frac{d\eta(t)}{dt}$ is $2ab\pi$. Since $\frac{d\log|t|}{dt} = \frac{1}{t}$, by applying L'Hospital's rule we get the result.

Write $f,u,v$ for $f(x,y),u(x,y),v(x,y)$ respectively.
If $d\log(u x^a)\wedge d\log(v y^b) = \omega \wedge \frac{dt}{t}$ where $\omega$ is a one form, since $d(\eta\{f_1,f_2\}) =d[\log |f_1| d\arg(f_2) -\log |f_2| d\arg(f_1)]= \im (d\log(f_1)\wedge d\log(f_2))$,
\begin{equation*}
\lim_{t\rightarrow 0} t\frac{d\eta(t)}{dt} = \lim_{t\rightarrow 0} \int_{\gamma_t} t \frac{d\eta(\{u x^a,v y^b\})}{dt} = \im\left(\lim_{t\rightarrow 0} \int_{\gamma_t} \omega\right).
\end{equation*}

Take $\log$ and differentiate $fxy=t$, we have
$$\frac{df}{f}+\frac{dx}{x}+\frac{dy}{y}=\frac{dt}{t}.$$
Now we have the following by substitute $dy$ by $dt$ and $dx$
\begin{eqnarray*}
d\log(u x^a)\wedge d\log(v y^b) &=& (d\log u + a d\log x)\wedge (d\log v + b d\log y) \\
&=& (a b + g y) \frac{dx}{x} \wedge \frac{dt}{t}
\end{eqnarray*}
where $g=g(x,y)$ is a holomorphic function of $x,y$ around $x=y=0$.

Since $\omega = (a b + g y) \frac{dx}{x}$, we have
\begin{equation*}
\im\left(\lim_{t\rightarrow 0} \int_{\gamma_t} \omega\right) =\im \left(\lim_{t\rightarrow 0} \int_{\gamma_t} (a b + g y) \frac{dx}{x}\right) = \im\left(\int_{\gamma_0} a b \frac{dx}{x}\right) = 2ab\pi
\end{equation*}
since $y=0$ on the loop $\gamma_0$.
\end{proof}


In order to calculate the limit of the regulator, we need a basis of $H_1(Y_t;\mathbb{Z})^{-}$ for $\abs{t}\rightarrow 0$.

Let $p_i$ be the point $(-a_i,1)$ on $Y_0$
This can be seen as all the intersection points of $x+a_i=0$ with $y-1=0$ (the thick points in Figure \ref{fig:badfiber}).
At any $p_i$, the surface is isomorphic to $fxy=1, f(p_i) \neq 0$ locally just by letting $x \rightarrow x-a_i$ and $y \rightarrow y+1$.
So as in Lemma \ref{lemma:limit} we can construct a loop $\gamma_i$ at $p_i$ which is continuous in the parameter $t$.



\begin{lemma}\label{lemma:basis}
These $g$ loops $\gamma_i, i = 1,\ldots,g$ give a basis of $H_1(Y_t;\mathbb{Z})^{-}$ for $\abs{t}\rightarrow 0$.
\end{lemma}
\begin{proof}
Note that on $Y_0$, the loop $\gamma_i$ is anti-invariant under the complex conjugation since the equations of the lines have real coefficients. The loop $\gamma_i$ is continuous in $t$, so on the fiber above $t$ close to 0 the loop is also anti-invariant under the complex conjugation, namely in $H_1(Y_t;\mathbb{Z})^{-}$.

Now we construct another set of loops on $Y$. On $Y_0$, we take the following $g$ rectangle loops
\begin{eqnarray*}
(0,-1) \rightarrow (-a_i,-1) \rightarrow (-a_i, 1) \rightarrow (0,1) \rightarrow (0,-1)
\end{eqnarray*}
which the $\rightarrow$ means the line between two points.

Let these loops vary continuously with $t$, we get another $g$ loops $\beta_i, i=1,\ldots,g$ on each fiber close to $t=0$.
Now $\beta_i$ and $\gamma_{j}$ can only intersect when $i=j$ and have intersection number $1$ when we choose the orientation of $\beta_i$ appropriately.

Now the intersection matrix of $\gamma_i, \beta_j, i,j = 1,\ldots,g$ looks like
\begin{equation*}
\left(
  \begin{array}{cc}
    0 & I_g \\
    -I_g & * \\
  \end{array}
\right)
\end{equation*}
which has determinant 1. So $\gamma_i, \beta_j, i,j = 1,\ldots,g$ give a basis of $H_1(Y_t;\mathbb{Z})$ for $\abs{t}\rightarrow 0$, consequently $\gamma_i, i =1,\ldots,g$ give a basis of $H_1(Y_t;\mathbb{Z})^{-}$.
\end{proof}

\begin{theorem}\label{thm:limit}
Let $C$ be defined by (\ref{eqn:curve}), we assume $t=1/\lambda, a_i \in \mathbb{R}^*$. If all $a_i$ are fixed and $t \rightarrow 0$ then the classes of $M_i$ in $K_2^{T}(C)/\torsion$ have regulator $R=R(t)$ satisfying
$$\lim_{t\rightarrow 0}\frac{R(t)}{(2\log \abs{t})^{g}}=1.$$
In particular, if $\lambda$ and $a_i$ are integers and $\lambda$ large enough, by Theorem \ref{thm:main} we have $g$ independent elements in $K_2(C;\mathbb{Z})$ as predicted by Beilinson's conjecture.
\end{theorem}


\begin{proof}
If we expand $\eta({M_{i}})$ by (\ref{eqn:element}) and apply Lemma \ref{lemma:limit}, it is easy to see that
\begin{equation*}
\frac{1}{2\pi}\lim_{t\rightarrow 0} \frac{\int_{\gamma_{i}}{\eta(M_{j})}}{\log \abs{t}} =
  \frac{1}{2\pi}\lim_{t\rightarrow 0} \frac{2\int_{\gamma_{i}}{\eta(\{x+a_j,y-1\})}}{\log \abs{t}} =
\begin{cases}
  2, & \text{if } i=j, \\
  0, & \text{otherwise},
\end{cases}
\end{equation*}



We now consider the $g \times g$ matrix
\begin{equation*}
M = \frac{1}{2\pi}
\left(\int_{\gamma_{i}}\eta(M_j)\right)_{1 \leqslant i,j \leqslant g},
\end{equation*}
since $\gamma_i,1\leqslant i\leqslant g$ give a basis of $H_1(X;\mathbb{Z})^{-}$ by Lemma \ref{lemma:basis}, we have
$$\lim_{t\rightarrow 0}\frac{R(t)}{(2\log \abs{t})^{g}}=
\lim_{t\rightarrow 0}\frac{\abs{\det(M)}}{(2\log \abs{t})^{g}}=1.$$
\end{proof}


\begin{remark}
The curve defined by (\ref{eqn:curve}) is isomorphic by replacing $(x,y)$ with $(\mu x , \mu^{g+1} y )$ to the curve defined by
\begin{equation} \label{eqn:curvetransform}
y(y+\lambda' \prod_{i=1}^g(a_i' x+1))=x^{2g+2}, \\
\end{equation}
 where $a_i'=\mu a_i, \lambda' = \lambda / \mu^{g+1}$. There are classes in $K_2^T(C)$ for this curve corresponding to $M_i$. So by using equation (\ref{eqn:curvetransform}) with $\mu = \abs{\lambda}^{1/(g+1)}$ we could assume that $\lambda = \pm 1$ and $a_i,b_i$ fixed. If we then let $\mu$ go to infinity we get a limit formula for the regulator $R(\mu)$ of the new family with parameter $\mu$.

\end{remark} 